在 QML 中访问 c​​pp 结构的最佳方法 [英] Best way to access a cpp structure in QML

问题描述

我需要在 cpp 和 QML 之间传递结构.如果我使用 property 我应该创建一个单独的集合并获取函数，我的结构包含至少 5 个成员，所以我觉得对所有这些成员使用集合和获取并不好.以下是我正在尝试做的一个例子:

I need to pass structures between cpp and QML. If i use property i should create an individual set and get functions, My structure contains minimum 5 members so i felt it's not good to use set and get for all those members. Following is an example of what i am trying to do :

MyClass.h

#include <QObject>
#include <QDebug>
using namespace std;

struct MyStruct {
Q_GADGET
int m_val;
QString m_name1;
QString m_name2;
QString m_name3;
QString m_name4;
Q_PROPERTY(int val MEMBER m_val)
Q_PROPERTY(QString name1 MEMBER m_name1)
Q_PROPERTY(QString name2 MEMBER m_name2)
Q_PROPERTY(QString name3 MEMBER m_name3)
Q_PROPERTY(QString name4 MEMBER m_name4)
};

class MyClass:public QObject
    {
        Q_OBJECT
    Q_PROPERTY(MyStruct mystr READ getMyStruct
                WRITE setMyStruct NOTIFY myStructChanged)

public:
    explicit MyClass(QObject *parent = nullptr);
    MyStruct strObj;

     // Edit: changed get function
     MyStruct getMyStruct() const
     {
         return strObj;
     }

// Edit: Added set function
     void setMyStruct(myStruct val)
        {
            mystr = val;
            emit myStructChanged();
        }

signals:
void myStructChanged();

}

ma​​in.cpp

#include <QGuiApplication>
#include <QQmlApplicationEngine>
#include <QQmlContext>
#include <QDebug>
#include <QObject>

#include "MyClass.h"

int main(int argc, char *argv[])
{
    QGuiApplication app(argc, argv);
    QQmlApplicationEngine engine;

    MyClass classObj;

    engine.rootContext()->setContextProperty("classObj",&classObj);

    engine.load(QUrl(QStringLiteral("qrc:/main.qml")));

    if (engine.rootObjects().isEmpty())
        return -1;

    return app.exec();
}

Main.qml

import QtQuick 2.6
import QtQuick.Controls 2.2
import QtQuick.Window 2.3

ApplicationWindow {

    id: applicationWindow

    visible: true
    width: 600
    height: 400
    title: qsTr("My App")

    MainForm{
        id : mainform

        Component.onCompleted: {
        console.log("name===="+classObj.mystr.name1)

        //EDIT added more code to explain the use case.
        classObj.myStr.name1 = "abc"  //Calls setter 
        classObj.mystr.name2 = "ans" // Calls setter 
        }
    }
}

如果我只打印 (classObj.myVariant) 我得到 QVariant(MyStruct) 但是当我尝试访问像 classObj.myVariant.name1 这样的任何参数时 我得到undefined"以及如何从 QML 设置变体?

If i print just (classObj.myVariant) i am getting QVariant(MyStruct) but when i tried to access any parameter like classObj.myVariant.name1 i am getting "undefined" and also how to set a variant from QML?

[UPDATE] - 还应将 MyStruct 添加到 Q_DECLARE_METATYPE，如下所示:Q_DECLARE_METATYPE(MyStruct)

[UPDATE] - Should also add MyStruct to Q_DECLARE_METATYPE as below: Q_DECLARE_METATYPE(MyStruct)

推荐答案

您需要元数据才能从 QML 访问 C++ 对象.

You need meta data to access C++ objects from QML.

对于非 QObject 派生，这是通过使用 Q_GADGET 宏并将成员公开为属性来实现的:

For non QObject derived, this is achieved by using the Q_GADGET macro, and exposing the members as properties:

struct MyStruct {
    Q_GADGET
    int m_val;
    QString m_name1;
    QString m_name2;
    QString m_name3;
    QString m_name4;
    Q_PROPERTY(int val MEMBER m_val)
    Q_PROPERTY(QString name1 MEMBER m_name1)
    Q_PROPERTY(QString name2 MEMBER m_name2)
    Q_PROPERTY(QString name3 MEMBER m_name3)
    Q_PROPERTY(QString name4 MEMBER m_name4)
};

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